Question

The maximum K.Es of the photoelectrons are $E_1$and $E_2$ with light of frequencies $v_1$and $v_2$ .The work function of the metal is

• A. $\frac{E_2{v}_1-E_1{v}_1}{v_1-v_2}$
• B. $\frac{E_2{v}_1-E_1{v}_1}{v_1+v_2}$
• C. $\frac{E_2{v}_2-E_1{v}_1}{v_1+v_2}$
• D. $\frac{E_2{v}_1-E_1{v}_2}{v_2-v_1}$

Answer 1

Answer: (D)

$\frac{E_2{v}_1-E_1{v}_2}{v_2-v_1}$

$h{v}_1=h{v}_0+E_1$
$h{v}_2=h{v}_0+E_2$
Sowing both $e{q}^{n}s\Rightarrow {\varphi }_0=\frac{E_2{v}_1-E_1{v}_2}{v_2-v_1}$